Answer Happy

A 6.234-mg sample produced 12.123 mg CO2 and 2.529 mg of H2O.
Find the wt% of C and H in the sample. (Answer: 53.07, 4.54%)
EXAMPLE Combustion Analysis Calculations A compound weighing 5.714 mg produced 14.414 mg of CO2 and 2.529 mg of H2O upon combustion. Find the wt% of C and H in the sample. g CO2 Solution One mole of CO2 contains 1 mol of carbon. Therefore, Moles of C in sample = moles of CO2 produced 14.414 x 10-3 = 3.275 X 10-4 mol 44.009 g/mol CO2 Mass of C in sample = (3.275 x 10-4 mol C)(12.010 6 g/mol C) = 3.934 mg 3.934 mg C wt%CE x 100 = 68.84% 5.714 mg sample One mole of H20 contains 2 mol of H. Therefore, Moles of H in sample = 2(moles of H2O produced) 2.529 X 10 g H2O = 2.808 x 10-4 mol 18.015 g/mol H2O = 2 Mass of H in sample = (2.808 x 10-4 mol H)(1.008 g/mol H) = 2.830 x 10-4 g 0.283 0 mg H wt% H = X 100 = 4.95% 5.714 mg sample TEST YOURSELF A 6.234-mg sample produced 12.123 mg CO2 and 2.529 mg of H,O. Find the wt% of C and H in the sample. (Answer: 53.07, 4.54%) Combustion Analysis Today13 Figure 27-8 shows an instrument that measures C, H, N, and S in a single operation. First, a ~2-mg sample is accurately weighed and sealed in a tin or silver capsule. The analyzer is swept with He gas that has been treated to remove traces of O2, H2O, and CO2. At the start of